Question

grayson owns a food truck that sells tacos and burritos. he only has enough supplies to make 110 tacos or burritos. he sells each taco for $3.25 and each burrito for $7. grayson must sell a minimum of $560 worth of tacos and burritos each day. if x represents the number of tacos sold and y represents the number of burritos sold, write and solve a system of inequalities graphically and determine one possible solution. inequality 1: y ≥ blank inequality 2: y ≥ blank

Explanation:

Step1: Define supply constraint

Total items made ≤ 110, so $x + y \leq 110$. Rearrange to solve for $y$:
$y \leq 110 - x$

Step2: Define revenue constraint

Total revenue ≥ $560$, so $3.25x + 7y \geq 560$. Rearrange to solve for $y$:
$7y \geq 560 - 3.25x$
$y \geq \frac{560 - 3.25x}{7}$
$y \geq 80 - \frac{13}{28}x$

Step3: Add non-negativity constraints

Since quantities can't be negative:
$x \geq 0$, $y \geq 0$

Step4: Identify feasible region

Graph the inequalities:

1. $y \leq 110 - x$ (line with intercepts (110,0) and (0,110), shade below)
2. $y \geq 80 - \frac{13}{28}x$ (line with intercepts $(\frac{560}{3.25} \approx 172.3,0)$ and (0,80), shade above)
3. $x \geq 0$, $y \geq 0$ (shade first quadrant)

The feasible region is the overlapping shaded area.

Step5: Find a valid solution

Pick a point in the feasible region, e.g., $x=40$:
Check $y \geq 80 - \frac{13}{28}(40) = 80 - \frac{130}{7} \approx 60.71$, so $y=60$ is too low, use $y=61$.
Verify supply: $40+61=101 \leq 110$
Verify revenue: $3.25(40)+7(61)=130+427=557 < 560$ (invalid). Try $y=62$:
Revenue: $130 + 7(62)=130+434=564 \geq 560$; Supply: $40+62=102 \leq 110$ (valid).

Answer:

Inequality 1: $y \leq 110 - x$
Inequality 2: $y \geq 80 - \frac{13}{28}x$
One possible solution: Sell 40 tacos and 62 burritos ($x=40, y=62$)